Solve (n+3)=(2n)^2+(-1+3)(2n)+(-1)(5)^1

$$( n + 3 ) = ( 2 n ) ^ { 2 } + ( - 1 + 3 ) ( 2 n ) + ( - 1 ) ( 5 ) ^ { 1 }$$

Solve for n

$$n = \frac{\sqrt{137} - 3}{8} \approx 1.088087489$$

$$n=\frac{-\sqrt{137}-3}{8}\approx -1.838087489$$

Steps Using the Quadratic Formula

Expand \left(2n\right)^{2}.

$$n+3=2^{2}n^{2}+\left(-1+3\right)\times 2n-5^{1}$$

Calculate 2 to the power of 2 and get 4.

$$n+3=4n^{2}+\left(-1+3\right)\times 2n-5^{1}$$

Add -1 and 3 to get 2.

$$n+3=4n^{2}+2\times 2n-5^{1}$$

Multiply 2 and 2 to get 4.

$$n+3=4n^{2}+4n-5^{1}$$

Calculate 5 to the power of 1 and get 5.

$$n+3=4n^{2}+4n-5$$

Subtract 4n^{2} from both sides.

$$n+3-4n^{2}=4n-5$$

Subtract 4n from both sides.

$$n+3-4n^{2}-4n=-5$$

Combine n and -4n to get -3n.

$$-3n+3-4n^{2}=-5$$

Add 5 to both sides.

$$-3n+3-4n^{2}+5=0$$

Add 3 and 5 to get 8.

$$-3n+8-4n^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-4n^{2}-3n+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, -3 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-4\right)\times 8}}{2\left(-4\right)}$$

Square -3.

$$n=\frac{-\left(-3\right)±\sqrt{9-4\left(-4\right)\times 8}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$n=\frac{-\left(-3\right)±\sqrt{9+16\times 8}}{2\left(-4\right)}$$

Multiply 16 times 8.

$$n=\frac{-\left(-3\right)±\sqrt{9+128}}{2\left(-4\right)}$$

Add 9 to 128.

$$n=\frac{-\left(-3\right)±\sqrt{137}}{2\left(-4\right)}$$

The opposite of -3 is 3.

$$n=\frac{3±\sqrt{137}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$n=\frac{3±\sqrt{137}}{-8}$$

Now solve the equation n=\frac{3±\sqrt{137}}{-8} when ± is plus. Add 3 to \sqrt{137}.

$$n=\frac{\sqrt{137}+3}{-8}$$

Divide 3+\sqrt{137} by -8.

$$n=\frac{-\sqrt{137}-3}{8}$$

Now solve the equation n=\frac{3±\sqrt{137}}{-8} when ± is minus. Subtract \sqrt{137} from 3.

$$n=\frac{3-\sqrt{137}}{-8}$$

Divide 3-\sqrt{137} by -8.

$$n=\frac{\sqrt{137}-3}{8}$$

The equation is now solved.

$$n=\frac{-\sqrt{137}-3}{8} n=\frac{\sqrt{137}-3}{8}$$

Steps for Completing the Square

Expand \left(2n\right)^{2}.

$$n+3=2^{2}n^{2}+\left(-1+3\right)\times 2n-5^{1}$$

Calculate 2 to the power of 2 and get 4.

$$n+3=4n^{2}+\left(-1+3\right)\times 2n-5^{1}$$

Add -1 and 3 to get 2.

$$n+3=4n^{2}+2\times 2n-5^{1}$$

Multiply 2 and 2 to get 4.

$$n+3=4n^{2}+4n-5^{1}$$

Calculate 5 to the power of 1 and get 5.

$$n+3=4n^{2}+4n-5$$

Subtract 4n^{2} from both sides.

$$n+3-4n^{2}=4n-5$$

Subtract 4n from both sides.

$$n+3-4n^{2}-4n=-5$$

Combine n and -4n to get -3n.

$$-3n+3-4n^{2}=-5$$

Subtract 3 from both sides.

$$-3n-4n^{2}=-5-3$$

Subtract 3 from -5 to get -8.

$$-3n-4n^{2}=-8$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-4n^{2}-3n=-8$$

Divide both sides by -4.

$$\frac{-4n^{2}-3n}{-4}=-\frac{8}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$n^{2}+\left(-\frac{3}{-4}\right)n=-\frac{8}{-4}$$

Divide -3 by -4.

$$n^{2}+\frac{3}{4}n=-\frac{8}{-4}$$

Divide -8 by -4.

$$n^{2}+\frac{3}{4}n=2$$

Divide \frac{3}{4}, the coefficient of the x term, by 2 to get \frac{3}{8}. Then add the square of \frac{3}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}+\frac{3}{4}n+\left(\frac{3}{8}\right)^{2}=2+\left(\frac{3}{8}\right)^{2}$$

Square \frac{3}{8} by squaring both the numerator and the denominator of the fraction.

$$n^{2}+\frac{3}{4}n+\frac{9}{64}=2+\frac{9}{64}$$

Add 2 to \frac{9}{64}.

$$n^{2}+\frac{3}{4}n+\frac{9}{64}=\frac{137}{64}$$

Factor n^{2}+\frac{3}{4}n+\frac{9}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n+\frac{3}{8}\right)^{2}=\frac{137}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+\frac{3}{8}\right)^{2}}=\sqrt{\frac{137}{64}}$$

Simplify.

$$n+\frac{3}{8}=\frac{\sqrt{137}}{8} n+\frac{3}{8}=-\frac{\sqrt{137}}{8}$$

Subtract \frac{3}{8} from both sides of the equation.

$$n=\frac{\sqrt{137}-3}{8} n=\frac{-\sqrt{137}-3}{8}$$